模板可能不是“虚拟"的 [英] templates may not be ‘virtual’

问题描述

给出以下代码，编译器将显示一条消息，指出该error: templates may not be ‘virtual’.有人对如何解决该错误有建议吗?

Given the code below, the compiler is showing a message pointing that error: templates may not be ‘virtual’. Does anyone have a suggestion on how to solve the bug?

template < class FOO_TYPE>
class CFoo{
    public:
        ...
        template < class BAR_TYPE >
        virtual void doSomething( const CBar<BAR_TYPE> &); // here's the error
        ...
        virtual ~CFoo();
    protected:
        MyClass < FOO_TYPE > * m_pClass;
};

template < class FOO_TYPE >
template < class BAR_TYPE >
void CFoo<FOO_TYPE>::doSomething( const CBar<BAR_TYPE> & refBar ){
    ...
}

推荐答案

了解为什么这是非法的最简单的原因是考虑使用vtable.当然，那只是一种常见的实现，而其他的实现也是允许的.但是C ++中的所有virtual函数都是经过设计的，因此可以使用vtable来实现.

The easiest reason to see why this is illegal is by considering the vtable. Sure, that's just one common implementation, and others are allowed. But all virtual functions in C++ are designed such that they can be implemented with a vtable.

现在，在CFoo<int>的vtable中有多少个条目?是否有doSomething<float>的条目?还有doSomething<float*>?还有doSomething<float**>吗?诸如此类的模板允许生成无限的功能集.通常这没问题，因为您只使用有限子集，但是对于虚函数，该子集是未知的，因此，vtable必须是无限的.

Now, how many entries are there in the vtable of CFoo<int> ? Is there an entry for doSomething<float> ? And doSomething<float*>? And doSomething<float**> ? Templates such as these allow an infinite set of functions to be generated. Usually that's no problem, as you use only a finite subset, but for virtual functions this subset isn't known, and therefore the vtable would need to be infinite.

现在，您可能真的只希望vtable中有一个条目.在这种情况下，您可以将其编写如下:

Now, it's possible that you really wanted only a single entry in the vtable. In that case, you'd write it as follows:

template < class FOO_TYPE, class BAR_TYPE>
class CFoo{
    public:
        ...
        virtual void doSomething( const CBar<BAR_TYPE> &); // now OK.
        ...
        virtual ~CFoo();
    protected:
        MyClass < FOO_TYPE > * m_pClass;
};

这意味着CFoo<int, float>的vtable将具有一个条目，对于doSomething(float const&).

This means that the vtable for CFoo<int, float> will have one entry, for doSomething(float const&).

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